A particle has a charge of q = 5e, where e is the charge on an electron.(a) Determine the electric potential (in V) due to the charge at a distance r = 0.260 cm from the charge.V(b) Determine the electric potential difference (in V) between a point that is 5r away and this point, that is V(5r) – V(r).Vraway and this point, that is V5) -(c) Determine the electric potential difference (in V) between a point that isV(r).V(d) How would the answers change if the electrons are replaced by protons?The sign of answer (a) would change.The sign of answer (b) would change.The sign of answer (c) would change.The sign of all answers would change.All answers would remain the same.

a) Given data The charge is given as q = 5e. The distance of the point from the charge is given as…

Answered: A particle has a charge of q = 5e,…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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A particle has a charge of q = 5e, where e is the charge on an electron.
(a) Determine the electric potential (in V) due to the charge at a distance r = 0.260 cm from the charge.
V
(b) Determine the electric potential difference (in V) between a point that is 5r away and this point, that is V(5r) – V(r).
V
r
away and this point, that is V
5
) -
(c) Determine the electric potential difference (in V) between a point that is
V(r).
V
(d) How would the answers change if the electrons are replaced by protons?
The sign of answer (a) would change.
The sign of answer (b) would change.
The sign of answer (c) would change.
The sign of all answers would change.
All answers would remain the same.

Expert Solution

Step 1

a) Given data

The charge is given as q = 5e.
The distance of the point from the charge is given as r = 0.260 cm.

The electric potential is calculated as,

$V (r) = \frac{k q}{r}$

Here, k is the coulomb's constant.

On putting the values as,

$V (r) = \frac{k q}{r} V (r) = \frac{9 \times 10^{9} {Nm}^{2} C^{- 2} \times (5 e)}{0.260 \times 10^{- 2} m} (1 e = - 1.6 \times 10^{- 19} C) V (r) = \frac{9 \times 10^{9} {Nm}^{2} C^{- 2} \times (5 \times - 1.6 \times 10^{- 19} C)}{0.260 \times 10^{- 2} m} V (r) = - 2.77 \times 10^{- 6} V$

Thus, the potential at a distance 0.260 cm far is -2.77 x 10^-6 V.

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